codeforces

Spreadsheets

time limit per test

10 seconds

memory limit per test

64 megabytes

input

standard input

output

standard output

In the popular spreadsheets systems (for example, in Excel) the following numeration of columns is used. The first column has number A, the second — number B, etc. till column 26 that is marked by Z. Then there are two-letter numbers: column 27 has number AA, 28 — AB, column 52 is marked by AZ. After ZZ there follow three-letter numbers, etc.

The rows are marked by integer numbers starting with 1. The cell name is the concatenation of the column and the row numbers. For example, BC23 is the name for the cell that is in column 55, row 23.

Sometimes another numeration system is used: RXCY, where X and Y are integer numbers, showing the column and the row numbers respectfully. For instance, R23C55 is the cell from the previous example.

Your task is to write a program that reads the given sequence of cell coordinates and produce each item written according to the rules of another numeration system.

Input

The first line of the input contains integer number n (1 ≤ n ≤ 105), the number of coordinates in the test. Then there follow n lines, each of them contains coordinates. All the coordinates are correct, there are no cells with the column and/or the row numbers larger than 106 .

Output

Write n lines, each line should contain a cell coordinates in the other numeration system.

Examples

input

2R23C55BC23

output

BC23R23C55

题意：

将R23C55（第23行55列）转换成另一种表达形式BC23，即将行用字母表示，A到Z（26位），下一位为AA（27），再例如BZ（2*26+26=78）。

以及将形如BC23转换成R23C55。

思路：

模拟一发即可。要注意判断到底是Excel转换成RXCY还是RXCY转换成Excel，不可以单纯的判断首字母为R，第二个字母为数字，就断定为是将RXCY转换成Excel格式。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <stack>using namespace std;const int MAXN = 1e6 + 10;int len;char str[MAXN];void one(){int pos, a=0, b=0;for(int i=0; i<len; ++i){if(str[i]=='C') {pos = i;break;}}for(int i=1; i<pos; ++i){a = a*10+(str[i]-'0');}for(int i=pos+1; i<len; ++i){b = b*10+(str[i]-'0');}stack<char> s;while(b>0){    if(b%26==0){            s.push('Z');            b -= 26;    }        else            s.push('A'-1+b%26);        b/=26;}while(!s.empty()){        printf("%c", s.top());        s.pop();}printf("%d\n", a);}void two(){int pos;for(int i=0; i<len; ++i){if(isdigit(str[i])){pos = i; break;}}int num = 0;for(int i=0; i<pos-1; ++i){num = num*26 + (str[i]-'A'+1)*26;}num += str[pos-1]-'A';printf("R");for(int i=pos; i<len; ++i) printf("%c", str[i]);printf("C%d\n", num+1);}int main(){int n;scanf("%d", &n);while(n--){        getchar();scanf("%s", str);len = strlen(str);bool aa = false;bool bb = false;for(int i=0; i<len; ++i){            if(isdigit(str[i])) aa = true;            if(aa && isalpha(str[i])){                bb = true; break;            }}if(bb)one();elsetwo();}return 0;}